- Try more complex generator
- ...so simpler the discriminator
- A "stronger" dataset: the style of the distribution should be more highlighted
Notes about training GAN
G' and G''
Exercises 7.41 and 8.14 from Rubinstein's book Polymer Physics.
From Boltzmann superposition principle, we have
$$\sigma(t)=\int_{-\infty}^t G(t-t^\prime)\dot{\gamma}(t^\prime)\mathrm{d}t^\prime$$
$$ \gamma(t) = \gamma_0 \exp(I \omega t)$$
$$\begin{align} \sigma(t)&=I\omega \gamma_0 \int_{-\infty}^t G(t-t^\prime)\exp(I \omega t^\prime)\mathrm{d}t^\prime\\ &=I \omega \gamma_0\exp(I \omega t)\int_0^\infty G(\tau)\exp(-I\omega \tau)\mathrm{d}\tau \\ &:=\gamma(t)I \omega \hat{G}(\omega)\end{align}$$
We therefore have
$$I\omega\hat{G}=I\omega \mathcal{F}^{\cos}\lbrace G\rbrace+\omega \mathcal{F}^{\sin}\lbrace G\rbrace\equiv I G^{\prime\prime}+G^\prime$$
$\mathcal{F}^{\cos}$ and $\mathcal{F}^{\sin}$ represent Fourier cosine and sine transform. Take $\mathcal{F}^{\cos}(1)=\delta(\omega)$ and $\mathcal{F}^{\sin}(1)=1/\omega$ (note the normalization factor of $\mathcal{F}$), let $G(t)=G_{eq}+G_1(t)$:
$$\begin{align} G^\prime &= \omega\mathcal{F}^{\sin}(G_1+G_{eq}) = G_{eq}+\omega\mathcal{F}^{\sin}(G_1(t))\\G^{\prime\prime}&=\omega\mathcal{F}^{\cos}(G_1+G_{eq})=\omega\mathcal{F}^{\cos}(G_1(t))+\omega\delta(\omega)=\omega\mathcal{F}^{\cos}(G_1(t))\end{align}$$
$G'$ and $G''$ are defined as the Fourier sine and cosine transform of stress relaxation function $G(t)$:
$$ \begin{align} G^\prime &:= \omega\int_0^\infty G(t) \sin\omega t\mathrm{d}t\\ G^{\prime\prime} &:= \omega\int_0^\infty G(t) \cos\omega t\mathrm{d}t \end{align}$$
Let $G(t)$ be
$$G(t)\sim \left(\frac{t}{\tau_0}\right)^{-\frac{1}{2}} \exp\left(-\frac{t}{\tau_R}\right)$$
G' i.e., the cosine transform is thus
$$G^{\prime\prime}/\omega=\int_0^\infty \mathrm{d}t \left(\frac{t}{\tau_0}\right)^{-\frac{1}{2}} \exp\left(-\frac{t}{\tau_R}\right)\cos\omega t$$
Since $\tau_R = N^2\tau_0$, we have (let ${t^\prime}^2\tau_R = t$)
$$\begin{align} G^{\prime\prime}/\omega &= 2\frac{\tau_R}{N} \int_0^\infty \mathrm{d}t^\prime \exp\left(-{t^\prime}^2\right)\cos\omega\tau_R {t^\prime}^2\\ &=2\frac{\tau_R}{N} \mathrm{Re} \int_0^\infty \mathrm{d}t^\prime \exp\left(-{t^\prime}^2\right)\exp\left(I \omega \tau_R {t^\prime}^2\right)\\ &=2\frac{\tau_R}{N} \mathrm{Re} \int_0^\infty \mathrm{d}t^\prime \exp\left(-(1-I \omega \tau_R){t^\prime}^2\right) \end{align} $$
Since $t^2=(-t)^2$, extending the range of last integral to $(-\infty, \infty)$ yields a Gaussian integral with complex argument, the result is straightforward:
$$ \begin{align}G^{\prime\prime}/\omega &=\frac{\tau_R}{N}\mathrm{Re}\sqrt{\frac{\pi}{1-I \omega \tau_R}}\\ &=\frac{\tau_R}{N}\mathrm{Re}\sqrt{\pi}\frac{\sqrt{1+I\omega\tau_R}}{1+(\omega\tau_R)^2}\\ &=\frac{\tau_R}{N}\mathrm{Re}\sqrt{\pi}\frac{1}{1+(\omega\tau_R)^2}\sqrt{\sqrt{1+(\omega\tau_R)^2}}\times&\\ &\left(\cos\left(\frac{1}{2}\arctan\omega\tau_R\right)+I \sin\left(\frac{1}{2}\arctan\omega\tau_R\right)\right)\\ &= \frac{\tau_R}{N}\sqrt{\pi}\frac{1}{1+(\omega\tau_R)^2}\sqrt{\sqrt{1+(\omega\tau_R)^2}}\cos\left(\frac{1}{2}\arctan\omega\tau_R\right)\end{align} $$ With $\cos(\arctan(x)) = \frac{1}{1+x^2}$ and $2\cos(x/2)^2-1=\cos(x)$, we finally have $$G^{\prime\prime} = \omega\tau_R\sqrt{\frac{\pi}{2}} \sqrt{\frac{\sqrt{1+(\omega\tau_R)^2}+1}{1+(\omega\tau_R)^2}}$$ The factor $\frac{\pi}{2}$ comes from the definition of Fourier cosine transform.
P.S. It is more convinient (non-restrictively) to treat the integral $$t^{-\alpha} \exp(-t) \exp(I w t)$$ with $0\lt \alpha \lt 1$ as Gamma function with (complex) coefficient in $\exp$: $$\int_0^\infty \mathrm{d}t t^{-\alpha}\exp(-(1-I \omega) t) = (1-I\omega t)^{\alpha-1} \Gamma(1-\alpha)$$ We may use Hypergeometric function $_pF_q$ for more genernal cases, i.e., $t^{-\alpha}\exp(-t^\beta)\exp(I w t)$
$$ \begin{align}G^{\prime\prime}/\omega &=\frac{\tau_R}{N}\mathrm{Re}\sqrt{\frac{\pi}{1-I \omega \tau_R}}\\ &=\frac{\tau_R}{N}\mathrm{Re}\sqrt{\pi}\frac{\sqrt{1+I\omega\tau_R}}{1+(\omega\tau_R)^2}\\ &=\frac{\tau_R}{N}\mathrm{Re}\sqrt{\pi}\frac{1}{1+(\omega\tau_R)^2}\sqrt{\sqrt{1+(\omega\tau_R)^2}}\times&\\ &\left(\cos\left(\frac{1}{2}\arctan\omega\tau_R\right)+I \sin\left(\frac{1}{2}\arctan\omega\tau_R\right)\right)\\ &= \frac{\tau_R}{N}\sqrt{\pi}\frac{1}{1+(\omega\tau_R)^2}\sqrt{\sqrt{1+(\omega\tau_R)^2}}\cos\left(\frac{1}{2}\arctan\omega\tau_R\right)\end{align} $$ With $\cos(\arctan(x)) = \frac{1}{1+x^2}$ and $2\cos(x/2)^2-1=\cos(x)$, we finally have $$G^{\prime\prime} = \omega\tau_R\sqrt{\frac{\pi}{2}} \sqrt{\frac{\sqrt{1+(\omega\tau_R)^2}+1}{1+(\omega\tau_R)^2}}$$ The factor $\frac{\pi}{2}$ comes from the definition of Fourier cosine transform.
P.S. It is more convinient (non-restrictively) to treat the integral $$t^{-\alpha} \exp(-t) \exp(I w t)$$ with $0\lt \alpha \lt 1$ as Gamma function with (complex) coefficient in $\exp$: $$\int_0^\infty \mathrm{d}t t^{-\alpha}\exp(-(1-I \omega) t) = (1-I\omega t)^{\alpha-1} \Gamma(1-\alpha)$$ We may use Hypergeometric function $_pF_q$ for more genernal cases, i.e., $t^{-\alpha}\exp(-t^\beta)\exp(I w t)$
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